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3 years ago

List THREE examples of where you would see ADDING or SUBTRACTING DECIMALS in the real world?

Mathematics

1 answer:

Blababa [14]3 years ago

6 0

1 Shopping.

2 Jobs.

3 Eating out.

Explanation:

1 When you go shopping you will need to add up your items too see the pricing.

2 Many jobs such as banking you need to know your adding subtracting and decimals.

3 Eating out you need to know how much the price is and how much to tip.

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A sample of 64 was taken from a population, with the mean of the sample being 119 and the standard deviation of the sample being

Fofino [41]

We have the sample size, sample mean and the sample standard deviation. Since the population standard deviation is not know, we will use t-distribution to find the confidence interval.

The critical t value for 95% confidence interval and 63 degrees of freedom is 1.998.

The 95% confidence for the population mean will be:

$(119-1.998 *\frac{16}{ \sqrt{64} } ,119+1.998 *\frac{16}{ \sqrt{64} }) \\ \\ 
=(115.004, 122.996) \\ \\=(115,123)$

Thus, the 95% confidence interval for the population mean will be (115,123)

So, option A is the correct answer

6 0

3 years ago

Hi guys, help with these two questions please! Will mark as brainliest!!

Colt1911 [192]

Step-by-step explanation:

Answer is attached....of both the parts.

Hope it helps:)

7 0

3 years ago

In a simple aquatic food chain, plants are eaten by small herbivorous fish. These small fish are eaten by larger fish. Which of

Romashka-Z-Leto [24]

It is secondary consumer because primary consumer is a animal that is a herbavorie (cant spell) and secondary eats that herbavor, and in this case the fish who ate the plant gets eaten by the bigger carnavore fish
IT IS C

7 0

3 years ago

Help pleaseeeeeeeeee

ad-work [718]

Answer:

non of these choices I guess

4 0

2 years ago

We have two fair three-sided dice, indexed by i = 1, 2. Each die has sides labeled 1, 2, and 3. We roll the two dice independent

Bogdan [553]

Answer:

(a) P(X = 0) = 1/3

(b) P(X = 1) = 2/9

(d) P(X = 3) = 0

(a) P(Y = 0) = 0

(b) P(Y = 1) = 1/3

Step-by-step explanation:

Given:

- Two 3-sided fair die.

- Random Variable X_1 denotes the number you get for rolling 1st die.

- Random Variable X_2 denotes the number you get for rolling 2nd die.

- Random Variable X = X_2 - X_1.

Solution:

- First we will develop a probability distribution of X such that it is defined by the difference of second and first roll of die.

- Possible outcomes of X : { - 2 , -1 , 0 ,1 , 2 }

- The corresponding probabilities for each outcome are:

( X = -2 ): { X_2 = 1 , X_1 = 3 }

P ( X = -2 ): P ( X_2 = 1 ) * P ( X_1 = 3 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

( X = -1 ): { X_2 = 1 , X_1 = 2 } + { X_2 = 2 , X_1 = 3 }

P ( X = -1 ): P ( X_2 = 1 ) * P ( X_1 = 3 ) + P ( X_2 = 2 ) * P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 0 ): { X_2 = 1 , X_1 = 1 } + { X_2 = 2 , X_1 = 2 } + { X_2 = 3 , X_1 = 3 }

P ( X = -1 ):P ( X_2 = 1 )*P ( X_1 = 1 )+P( X_2 = 2 )*P ( X_1 = 2)+P( X_2 = 3 )*P ( X_1 = 3)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 3 / 9 ) = ( 1 / 3 )

( X = 1 ): { X_2 = 2 , X_1 = 1 } + { X_2 = 3 , X_1 = 2 }

P ( X = 1 ): P ( X_2 = 2 ) * P ( X_1 = 1 ) + P ( X_2 = 3 ) * P ( X_1 = 2)

: ( 1 / 3 ) * ( 1 / 3 ) + ( 1 / 3 ) * ( 1 / 3 )

: ( 2 / 9 )

( X = 2 ): { X_2 = 1 , X_1 = 3 }

P ( X = 2 ): P ( X_2 = 3 ) * P ( X_1 = 1 )

: ( 1 / 3 ) * ( 1 / 3 )

: ( 1 / 9 )

- The distribution Y = X_2,

P(Y=0) = 0

P(Y=1) = 1/3

P(Y=2) = 1/ 3

- The probability for each number of 3 sided die is same = 1 / 3.

7 0

3 years ago